I found a definition: Chartjunk<span> refers to all visual elements in </span>charts<span> and graphs that are not necessary to comprehend the information represented on the graph, or that distract the viewer from this information.</span>
Answer:
This is a conceptual problem so I will try my best to explain the impossible scenario. First of all the two dust particles ara virtually exempt from any external forces and at rest with respect to each other. This could theoretically happen even if it's difficult for that to happen. The problem is that each of the particles have an electric charge which are equal in magnitude and sign. Thus each particle should feel the presence of the other via a force. The forces felt by the particles are equal and opposite facing away from each other so both charges have a net acceleration according to Newton's second law because of the presence of a force in each particle:

Having seen Newton's second law it should be clear that the particles are actually moving away from each other and will not remain at rest with respect to each other. This is in contradiction with the last statement in the problem.
Explanation:
Radius of a charged particle is given by
r=mv / Bq
= k/ q
where k = m v / B is a constant.
i.e. more is the magnitude of charge, less is the radius.
(inversely proportional)
From the diagram r_3 > r_2 > r_1 (more the curvature, less is the radius)
( although drawing is not given i am assuming the above order, however, one can change the order as per the diagram. The concept used remains the same)
therefore, q_1 > q_2 > q_3
.
Answer:

(Downwards)
(Towards Left)
Explanation:
As we know that beam is in equilibrium
So here we can use torque balance as well as force balance for the beam
Now by torque balance equation at the pivot we can say

As we know that
mg = 1.40 kN
F = 5 kN
so we will have


Now force balance in vertical direction


(Downwards)
Force balance in horizontal direction


(Towards Left)
#1
Volume of lead = 100 cm^3
density of lead = 11.34 g/cm^3
mass of the lead piece = density * volume


so its weight in air will be given as

now the buoyant force on the lead is given by


now as we know that


so by solving it we got
V = 11.22 cm^3
(ii) this volume of water will weigh same as the buoyant force so it is 0.11 N
(iii) Buoyant force = 0.11 N
(iv)since the density of lead block is more than density of water so it will sink inside the water
#2
buoyant force on the lead block is balancing the weight of it




(ii) So this volume of mercury will weigh same as buoyant force and since block is floating here inside mercury so it is same as its weight = 11.11 N
(iii) Buoyant force = 11.11 N
(iv) since the density of lead is less than the density of mercury so it will float inside mercury
#3
Yes, if object density is less than the density of liquid then it will float otherwise it will sink inside the liquid